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这是美国康涅狄格大学微积分课程的代写成功案例。
Math1131课程简介
Course communication: Much of the communication for this course will occur online, via HuskyCT, email, and/or this page. Please monitor this page, course HuskyCT announcements, and your UConn email regularly for updates related to MATH 1131Q. However, you are also expected to be in class every day to receive important communications there, and to contact a classmate, or your instructor or TA, if you miss class to make sure you have not missed important course information.
Prerequisites
Time Zone: All times for due dates, online and in-person classes, office hours, Supplemental Instruction, and Q Center hours in this course are given in US Eastern Time.
Description: This course is an introduction to differential and integral calculus, which is the mathematical language used in any science concerned with dynamically changing quantities. The main topics it covers are limits, derivatives, integrals, the Fundamental Theorem of Calculus, and some basic applications of these ideas.
Prerequisites: A qualifying score of 22 on the mathematics placement exam (MPE). Students who fail to achieve this minimum score are required to spend time on the preparatory and learning modules before re-taking the MPE, or to register for a lower level Mathematics course. Not open for credit to students who have passed MATH 1132Q or 1152Q.
Math1131 Calculus HELP（EXAM HELP， ONLINE TUTOR）
Suppose $\left(a_n\right)$ converges to $a$. Then, by Definition 1.1.2, taking $\varepsilon=10^{-k}$ for some $k \in \mathbb{N}$, there exists some $N_k \in \mathbb{N}$ such that
$$
\left|a_n-a\right|<10^{-k} \quad \forall n \geq N_k .
$$
In other words:
If $n \geq N_k$ and if $a_n$ and $a$ are expressed in decimal expansion, then the first $k$ decimal places of $a_n$ and $a$ are the same
Note that, in the definition of convergence, different $\varepsilon$ can result in different $N$, i.e., the number $N$ may vary as $\varepsilon$ varies. This fact will be revealed in the following examples.
The following theorem is useful for showing convergence of certain sequences using the convergence of some other sequences.
Theorem 1.1.3 Suppose $a_n \rightarrow a$ and $b_n \rightarrow$ bas $n \rightarrow \infty$. Then the following results hold.
(i) $a_n+b_n \rightarrow a+b$ as $n \rightarrow \infty$.
(ii) $c a_n \rightarrow$ ca as $n \rightarrow \infty$ for any real number $c$.
(iii) If $a_n \leq b_n$ for all $n \in \mathbb{N}$, then $a \leq b$.
(iv) (Sandwich theorem) If $a_n \leq c_n \leq b_n$ for all $n \in \mathbb{N}$, and if $a=b$, then $c_n \rightarrow a$ as $n \rightarrow \infty$.
Show that $|a|<1$ implies $a^n \rightarrow 0$.
For $a \geq 0$, prove the following. (i) $a^n \rightarrow 0 \Longleftrightarrow a<1$. (ii) $a^n \rightarrow \infty \Longleftrightarrow a>1$.
If $a_n \rightarrow a$ and $a \neq 0$, then show that there exists $k \in \mathbb{N}$ such that $a_n \neq 0$ for all $n \geq k$.
Consider the sequence $\left(a_n\right)$ with $a_n=\left(1+\frac{1}{n}\right)^{1 / n}, n \in \mathbb{N}$. Then show that $\lim {n \rightarrow \infty} a_n=1$. Hint: Observe that $1 \leq a_n \leq(1+1 / n)$ for all $n \in \mathbb{N}$.
Consider the sequence $\left(a_n\right)$ with $a_n=\frac{1}{n^k}, n \in \mathbb{N}$. Then show that for any given $k \in \mathbb{N}, \lim {n \rightarrow \infty} a_n=0$ (Fig. 1.5).
Hint: Observe that $1 \leq a_n \leq 1 / n$ for all $n \in \mathbb{N}$.
Establish the statement in the last paragraph of the above remark. $\triangleleft$
The following theorem gives a sufficient condition for certain number to be a limit of a given sequence.
Theorem 1.1.6 Let $\left(a_n\right)$ be a sequence such that
$$
\left|a_{n+1}-a\right| \leq r\left|a_n-a\right| \quad \forall n \in \mathbb{N}
$$
for some $a \in \mathbb{R}$ and for some $r$ with $0<r<1$. Then $a_n \rightarrow a$.
$\operatorname{Proof}$ For each $n \in \mathbb{N}$, we have
$$
\left|a_{n+1}-a\right| \leq r\left|a_n-a\right| \leq \cdots \leq r^n\left|a_1-a\right|
$$
Thus,
$$
0 \leq\left|a_{n+1}-a\right| \leq r^n\left|a_1-a\right| \quad \forall n \in \mathbb{N}
$$
Since $0<r<1$, by Theorem $1.1 .4, r^n \rightarrow 0$. Hence, by Sandwich theorem, $a_n \rightarrow a$.
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